Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $n = \dfrac{-3t + 30}{6t - 6} \div \dfrac{-8t + 80}{t^2 + 5t - 6} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{-3t + 30}{6t - 6} \times \dfrac{t^2 + 5t - 6}{-8t + 80} $ First factor the quadratic. $n = \dfrac{-3t + 30}{6t - 6} \times \dfrac{(t - 1)(t + 6)}{-8t + 80} $ Then factor out any other terms. $n = \dfrac{-3(t - 10)}{6(t - 1)} \times \dfrac{(t - 1)(t + 6)}{-8(t - 10)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ -3(t - 10) \times (t - 1)(t + 6) } { 6(t - 1) \times -8(t - 10) } $ $n = \dfrac{ -3(t - 10)(t - 1)(t + 6)}{ -48(t - 1)(t - 10)} $ Notice that $(t - 10)$ and $(t - 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ -3\cancel{(t - 10)}(t - 1)(t + 6)}{ -48\cancel{(t - 1)}(t - 10)} $ We are dividing by $t - 1$ , so $t - 1 \neq 0$ Therefore, $t \neq 1$ $n = \dfrac{ -3\cancel{(t - 10)}\cancel{(t - 1)}(t + 6)}{ -48\cancel{(t - 1)}\cancel{(t - 10)}} $ We are dividing by $t - 10$ , so $t - 10 \neq 0$ Therefore, $t \neq 10$ $n = \dfrac{-3(t + 6)}{-48} $ $n = \dfrac{t + 6}{16} ; \space t \neq 1 ; \space t \neq 10 $